I think there are 3 properties to check? Seeing if 0 "is an element of" H; Seeing if u + v "is an element of" H, and if u "is an element of H" then is cu "an element of H" for c "an element of" R.

November 30th 2006, 03:51 AM

TD!

That's right, H has to be non-empty (easy to check: 0 has to be an element) and you can combine the other two properties into one: it has to be closed under lineair combinations, i.e. if v and w (vectors) are solutions, then kv+lw (k,l scalars) has to be a solution too.

1) 0 is a solution, trivial. If (a,b,c,d) and (e,f,g,h) are solutions, is k(a,b,c,d)+l(e,f,h,h) a solution too? Check it :)

2) Try it.

December 3rd 2006, 09:39 PM

fifthrapiers

Hmm. Okay so there are 3 things you have to check for each. So I believe its something along check if the zero vector is an element of H, and it H is "closed" under adddition and multiplication.

I dont really get what the a + b + c + d = 0 is for... it seems obvious that H is a subspace for a...what is the basis for it? How do I show it?

#2: again, like #1, it seems obvious H is a subspace...same questions. A detailed answer will help me do some of the other homework problems.

Thanks

December 4th 2006, 06:20 AM

fifthrapiers

Tried doing some other of the others, though they're all like this and ask the same question. I thought it would be sufficient to prove if the 0 zero is an element of H only, but not according to my prof...

Anyone familiar with subspace?

December 4th 2006, 12:43 PM

TD!

Taking the two vectors I suggested, we want to know whether the following is in H too:

This is only true if ka+le + kb+lf + kc+lg + kd+lh = 0. But we can rewrite: